n^2-48=8n

Solution for n^2-48=8n equation:

n^2-48=8n

We move all terms to the left:

n^2-48-(8n)=0

a = 1; b = -8; c = -48;
Δ = b2-4ac
Δ = -82-4·1·(-48)
Δ = 256
The delta value is higher than zero, so the equation has two solutions
We use following formulas to calculate our solutions:
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}$

$\sqrt{\Delta}=\sqrt{256}=16$
$n_{1}=\frac{-b-\sqrt{\Delta}}{2a}=\frac{-(-8)-16}{2*1}=\frac{-8}{2}
=-4
$
$n_{2}=\frac{-b+\sqrt{\Delta}}{2a}=\frac{-(-8)+16}{2*1}=\frac{24}{2}
=12
$